Question: A circle centered at $O$ is circumscribed about $\triangle ABC$ as follows: [asy]
pair pA, pB, pC, pO;
pO = (0, 0);
pA = pO + dir(-20);
pB = pO + dir(90);
pC = pO + dir(190);
draw(pA--pB--pC--pA);
draw(pO--pA);
draw(pO--pB);
draw(pO--pC);
label("$O$", pO, S);
label("$110^\circ$", pO, NE);
label("$100^\circ$", pO, NW);
label("$A$", pA, SE);
label("$B$", pB, N);
label("$C$", pC, SW);
draw(circle(pO, 1));
[/asy] What is the measure of $\angle BAC$, in degrees?
Answer: We can see that $\angle AOC = 360^\circ - (110^\circ + 100^\circ) = 150^\circ.$ Now, $\triangle AOC$ and $\triangle AOB$ are both isosceles triangles. That means that $\angle OAC = \frac{1}{2} \cdot (180^\circ - 150^\circ) = 15^\circ$ and $\angle OAB = \frac{1}{2} \cdot (180^\circ - 110^\circ) = 35^\circ.$  Therefore, our answer is $\angle BAC = \angle OAB + \angle OAC = 15^\circ + 35^\circ = \boxed{50^\circ}.$